Partial-trace data processing for finite quantum relative entropy

Choose an orthonormal basis $(e_j{)}_{j<d}$ of $H_B$ using [F8], and set $\omega =\exp (2\pi i\mathrm{/}d)$. By [F10] and [F12], $\mathrm{\mid }\omega \mathrm{\mid }=1$ and ${\omega }^m=\exp (2\pi im\mathrm{/}d)$ for every integer $m$; by [F11], ${\omega }^m=1$ exactly when $d$ divides $m$. Thus $\omega$ has order $d$. Define $X{e}_j=e_{j+1<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>d}$ and $Z{e}_j={\omega }^j{e}_j$; these maps are unitary because they respectively permute the basis and multiply its vectors by unit-modulus scalars. If $d\nmid m$, then $q={\omega }^m\mathrm{\ne }1$, $q^d=1$, and $(1-q){\sum }_{b=0}^{d-1}{q}^b=1-q^d=0$, so $d^{-1}{\sum }_{b=0}^{d-1}{\omega }^{bm}=0$; if $d\mid m$, that average is $1$. Choose an orthonormal basis $(u_i)$ of $H_A$ by [F8]. For $0\le r,s<d$, define the linear block $R_{rs}:H_A\to H_A$ by $R_{rs}{u}_{i^{\mathrm{\prime }}}={\sum }_i⟨R(u_{i^{\mathrm{\prime }}}\otimes e_s),u_i\otimes e_r⟩u_i$, extending linearly. Expanding $R(u_{i^{\mathrm{\prime }}}\otimes e_s)$ in the product basis [F5] then gives $R={\sum }_{r,s}{R}_{rs}\otimes M_{rs}$, where $M_{rs}z:=⟨z,e_s⟩e_r$: both sides agree on every product basis vector. Direct application to basis vectors gives $Z^b{M}_{rs}(Z^b{)}^{\ast }={\omega }^{b(r-s)}{M}_{rs}$ and $X^a{M}_{rs}(X^a{)}^{\ast }=M_{r+a,s+a}$, with indices modulo $d$. For $0\le r,s<d$, $d\mid (r-s)$ exactly when $r=s$, so the proved phase average removes precisely the off-diagonal units. Also $d^{-1}{\sum }_{a=0}^{d-1}{M}_{r+a,r+a}=I_B\mathrm{/}d$. Therefore $\mathcal{T}(R):=d^{-2}{\sum }_{a,b}(I_A\otimes X^a{Z}^b)R(I_A\otimes X^a{Z}^b{)}^{\ast }={\sum }_r{R}_{rr}\otimes I_B\mathrm{/}d={\mathrm{Tr}}_B(R)\otimes I_B\mathrm{/}d$ by [F3]. For every $a,b$, the lifted map $I_A\otimes X^a{Z}^b$ sends $u_i\otimes e_j$ to ${\omega }^{bj}{u}_i\otimes e_{j+a}$, hence sends an orthonormal product basis to an orthonormal product basis; it is therefore unitary, including $d=1$. When $d=1$, $X=Z=I_B$ and there is only the diagonal unit, so the same formula holds directly.

For non-negative $C,D$ on $H_A$, their tensor products with $I_B\mathrm{/}d$ have supports $\mathrm{supp}(C)\otimes H_B$ and $\mathrm{supp}(D)\otimes H_B$ by the tensor-product eigenbases [F7]. Thus support inclusion holds for $(C,D)$ exactly when it holds for $(C\otimes I_B\mathrm{/}d,D\otimes I_B\mathrm{/}d)$. In the finite branch, each positive tensor eigenvalue has the form $\lambda \mathrm{/}d$, and [F13] gives $\ln (\lambda \mathrm{/}d)=\ln \lambda -\ln d$; the two logarithms therefore have the same additive term $-\ln d$, which cancels, while $\mathrm{tr}(I_B\mathrm{/}d)=1$; hence $D(C\otimes I_B\mathrm{/}d\mathrm{\parallel }D\otimes I_B\mathrm{/}d)=D(C\mathrm{\parallel }D)$. If inclusion fails both sides are $+\mathrm{\infty }$ by [F6].

Apply joint convexity [F1] to the finite average defining $\mathcal{T}$ and isometric invariance [F2] to each unitary conjugate. This gives $D(\mathcal{T}(A)\mathrm{\parallel }\mathcal{T}(B))\le D(A\mathrm{\parallel }B)$.

By [F4], $C={\mathrm{Tr}}_{B}A$ and $D={\mathrm{Tr}}_{B}B$ are non-negative. Substitute them into step 1.2, use step 1.1 to identify their twirls, and then apply step 2.1 to obtain $D({\mathrm{Tr}}_{B}A\mathrm{\parallel }{\mathrm{Tr}}_{B}B)\le D(A\mathrm{\parallel }B)$.